\(\int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1461]

Optimal result

Integrand size = 29, antiderivative size = 87 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 a^2 b \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 \cot (c+d x)}{d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+\frac {3 a b^2 \tan (c+d x)}{d} \]

[Out]

-3*a^2*b*arctanh(cos(d*x+c))/d-a^3*cot(d*x+c)/d+3*a^2*b*sec(d*x+c)/d+b^3*sec(d*x+c)/d+a^3*tan(d*x+c)/d+3*a*b^2
*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2991, 3852, 8, 2702, 327, 213, 2700, 14, 2686} \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3 \tan (c+d x)}{d}-\frac {a^3 \cot (c+d x)}{d}-\frac {3 a^2 b \text {arctanh}(\cos (c+d x))}{d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d} \]

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]

[Out]

(-3*a^2*b*ArcTanh[Cos[c + d*x]])/d - (a^3*Cot[c + d*x])/d + (3*a^2*b*Sec[c + d*x])/d + (b^3*Sec[c + d*x])/d +
(a^3*Tan[c + d*x])/d + (3*a*b^2*Tan[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2991

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (3 a b^2 \sec ^2(c+d x)+3 a^2 b \csc (c+d x) \sec ^2(c+d x)+a^3 \csc ^2(c+d x) \sec ^2(c+d x)+b^3 \sec (c+d x) \tan (c+d x)\right ) \, dx \\ & = a^3 \int \csc ^2(c+d x) \sec ^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \csc (c+d x) \sec ^2(c+d x) \, dx+\left (3 a b^2\right ) \int \sec ^2(c+d x) \, dx+b^3 \int \sec (c+d x) \tan (c+d x) \, dx \\ & = \frac {a^3 \text {Subst}\left (\int \frac {1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (3 a^2 b\right ) \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac {\left (3 a b^2\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}+\frac {b^3 \text {Subst}(\int 1 \, dx,x,\sec (c+d x))}{d} \\ & = \frac {3 a^2 b \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d}+\frac {3 a b^2 \tan (c+d x)}{d}+\frac {a^3 \text {Subst}\left (\int \left (1+\frac {1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (3 a^2 b\right ) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {3 a^2 b \text {arctanh}(\cos (c+d x))}{d}-\frac {a^3 \cot (c+d x)}{d}+\frac {3 a^2 b \sec (c+d x)}{d}+\frac {b^3 \sec (c+d x)}{d}+\frac {a^3 \tan (c+d x)}{d}+\frac {3 a b^2 \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.31 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\left (2 a^3+3 a b^2\right ) \cos (2 (c+d x))-2 b \left (3 a^2+b^2\right ) \sin (c+d x)-3 a b \left (b+a \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin (2 (c+d x))\right )\right )}{4 d} \]

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]

[Out]

-1/4*(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*Sec[c + d*x]*((2*a^3 + 3*a*b^2)*Cos[2*(c + d*x)] - 2*b*(3*a^2 + b^2)*S
in[c + d*x] - 3*a*b*(b + a*(-Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]])*Sin[2*(c + d*x)])))/d

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 a^{2} b \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a \,b^{2} \tan \left (d x +c \right )+\frac {b^{3}}{\cos \left (d x +c \right )}}{d}\)	\(91\)
default	\(\frac {a^{3} \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )+3 a^{2} b \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a \,b^{2} \tan \left (d x +c \right )+\frac {b^{3}}{\cos \left (d x +c \right )}}{d}\)	\(91\)
parallelrisch	\(\frac {6 b \,a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (-a^{3}-2 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{3} \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-12 a^{2} b -4 b^{3}}{2 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(113\)
risch	\(-\frac {2 i \left (3 i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-3 i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-i b^{3} {\mathrm e}^{i \left (d x +c \right )}-3 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 a^{3}+3 a \,b^{2}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(165\)
norman	\(\frac {\frac {a^{3}}{2 d}+\frac {a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {2 \left (3 a^{2} b +b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (6 a^{2} b +2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (9 a^{2} b +3 b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a \left (a^{2}+4 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 a \left (a^{2}+4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (7 a^{2}+18 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (7 a^{2}+18 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 b \left (3 a^{2}+b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 a^{2} b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(305\)

[In]

int(csc(d*x+c)^2*sec(d*x+c)^2*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(1/sin(d*x+c)/cos(d*x+c)-2*cot(d*x+c))+3*a^2*b*(1/cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+3*a*b^2*tan(d
*x+c)+b^3/cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.51 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 \, a^{2} b \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, a^{2} b \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a^{3} - 6 \, a b^{2} + 2 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(3*a^2*b*cos(d*x + c)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*a^2*b*cos(d*x + c)*log(-1/2*cos(d*x +
c) + 1/2)*sin(d*x + c) - 2*a^3 - 6*a*b^2 + 2*(2*a^3 + 3*a*b^2)*cos(d*x + c)^2 - 2*(3*a^2*b + b^3)*sin(d*x + c)
)/(d*cos(d*x + c)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**2*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, a^{2} b {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{3} {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + 6 \, a b^{2} \tan \left (d x + c\right ) + \frac {2 \, b^{3}}{\cos \left (d x + c\right )}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(3*a^2*b*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 2*a^3*(1/tan(d*x + c) - tan(d*
x + c)) + 6*a*b^2*tan(d*x + c) + 2*b^3/cos(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.70 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {6 \, a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {2 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(6*a^2*b*log(abs(tan(1/2*d*x + 1/2*c))) + a^3*tan(1/2*d*x + 1/2*c) - (2*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 5*a
^3*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 10*a^2*b*tan(1/2*d*x + 1/2*c) + 4*b^3*tan(1/2*d*
x + 1/2*c) - a^3)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)))/d

Mupad [B] (verification not implemented)

Time = 11.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.38 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2\,b+4\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^3+12\,a\,b^2\right )-a^3}{d\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {3\,a^2\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

[In]

int((a + b*sin(c + d*x))^3/(cos(c + d*x)^2*sin(c + d*x)^2),x)

[Out]

(a^3*tan(c/2 + (d*x)/2))/(2*d) + (tan(c/2 + (d*x)/2)*(12*a^2*b + 4*b^3) + tan(c/2 + (d*x)/2)^2*(12*a*b^2 + 5*a
^3) - a^3)/(d*(2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^3)) + (3*a^2*b*log(tan(c/2 + (d*x)/2)))/d